A crate is pulled right with a force of 82.0N to the left wtih a force of 115N upward with a force of 565N and downward with a force of 236N
Fine the net external force in the X direction
Find the net external force in the y direction
find the the magnitude and direction of the net external force on the crate
I'm lost
I think you meant -5.73 round it up call it -6. The annotated teachers version says the angle is 96 degrees.... is there some connection here where 96-6 gives you a right triangle? Maybe my thinking is wrong.
Oh no! Don't worry, I'm here to help you find your way back! Let's break it down:
To find the net external force in the x direction, we need to sum up all the forces acting horizontally. The only force acting horizontally is the force pulling the crate to the right with a magnitude of 82.0N. So, the net external force in the x direction is simply 82.0N.
Now, let's find the net external force in the y direction. We have two forces acting vertically: 115N upward and 236N downward. To calculate the net force in the y direction, we subtract the downward force from the upward force: 115N - 236N = -121N. So, the net external force in the y direction is -121N.
To find the magnitude and direction of the net external force on the crate, we can use the Pythagorean theorem. The magnitude of the net external force is the square root of the sum of the squares of the x and y components of the force:
Magnitude = sqrt((82.0N)^2 + (-121N)^2)
Calculating this gives us a magnitude of approximately 147.03N.
To determine the direction of the net external force, we can use trigonometry and find the angle theta:
Theta = arctan((net force in y direction)/(net force in x direction))
Plugging in the values, we get:
Theta = arctan((-121N)/(82.0N))
Calculating this gives us an angle of approximately -55.04 degrees.
So, the magnitude of the net external force is approximately 147.03N, and the direction is approximately -55.04 degrees.
To find the net external force, we need to consider the forces acting in the x and y directions separately and then combine them.
In the x direction:
- The force of 82.0N to the right is positive (since it is in the positive x-direction).
- The force of 115N to the left is negative (since it is in the negative x-direction).
To find the net external force in the x-direction, we need to add these forces:
Net external force (x-direction) = 82.0N - 115N = -33.0N (negative because the net force is in the negative x-direction)
In the y direction:
- The force of 565N upwards is positive (since it is in the positive y-direction).
- The force of 236N downwards is negative (since it is in the negative y-direction).
To find the net external force in the y-direction, we need to add these forces:
Net external force (y-direction) = 565N - 236N = 329N (positive because the net force is in the positive y-direction)
To find the magnitude and direction of the net external force on the crate, we can use the Pythagorean theorem:
Magnitude of net external force = sqrt((net external force in x-direction)^2 + (net external force in y-direction)^2)
Magnitude of net external force = sqrt((-33.0N)^2 + (329N)^2) ≈ 332.5N
The direction of the net external force can be found by calculating the angle using:
Angle = arctan(net external force in y-direction / net external force in x-direction)
Angle ≈ arctan(329N / -33.0N) ≈ -81.3° (negative because the angle is measured counterclockwise from the positive x-axis)
Therefore, the magnitude of the net external force on the crate is approximately 332.5N, and the direction is approximately -81.3° relative to the positive x-axis.
331N at 5.73
If I read this correctly, the net force to the right X direction is 82-115
and net upward of 565-236
Not to get the net magnitude, use Pythagorean theorm. For the angle, use the tangent function.